Limits question

**akhayoon** · April 21st 2008, 08:43 AM

Limit question on a midterm that I have and still haven't the slightest idea on how to complete

Find all Values of a and b such that $\lim{x}\rightarrow{0}\frac{3sin(2x)+ax^{2}-bx}{x^2}=1$

all I could figure out was that if you use L'Hopitale's rule you would get

$\lim{x}\rightarrow{0}\frac{6cos(2x)+2ax+b}{2x}=1$

but I have no idea how to use this to solve for any variables

thank you

**Moo** · April 21st 2008, 08:46 AM

Hello,

Why not using a second time l'Hôpital's rule ?

**akhayoon** · April 21st 2008, 08:54 AM

because it would be (0/0) ?

**Moo** · April 21st 2008, 08:57 AM

Hm ok, I just don't know correctly the conditions for applying this rule

**flyingsquirrel** · April 21st 2008, 09:00 AM

Hi

Another approach : $\lim_{x\to 0} \frac{\sin(2x)-\sin(0)}{2x-0}=\cos 0 = 1$
Hence, $\lim_{x \to 0} \frac{3}{x}\frac{\sin(2x)}{x}+a-\frac{b}{x}=\lim_{x \to 0} \frac{6}{x} \frac{\sin(2x)}{2x}+a-\frac{b}{x}=\lim_{x\to 0}\frac{6-b}{x}+a=1$
When is this possible ?

**Mathstud28** · April 21st 2008, 11:58 AM

Originally Posted by akhayoon

Limit question on a midterm that I have and still haven't the slightest idea on how to complete

Find all Values of a and b such that $\lim{x}\rightarrow{0}\frac{3sin(2x)+ax^{2}-bx}{x^2}=1$

all I could figure out was that if you use L'Hopitale's rule you would get

$\lim{x}\rightarrow{0}\frac{6cos(2x)+2ax+b}{2x}=1$

but I have no idea how to use this to solve for any variables

thank you

L'hopital it again if you wish...it works...but look...you have two equations with two variables...solve for them using simulataneous equations

**mr fantastic** · April 21st 2008, 04:05 PM

Originally Posted by akhayoon

Limit question on a midterm that I have and still haven't the slightest idea on how to complete

Find all Values of a and b such that $\lim{x}\rightarrow{0}\frac{3sin(2x)+ax^{2}-bx}{x^2}=1$

all I could figure out was that if you use L'Hopitale's rule you would get

$\lim{x}\rightarrow{0}\frac{6cos(2x)+2ax+b}{2x}=1$

but I have no idea how to use this to solve for any variables

thank you

Obviously you require another indeterminant form. This will happen if b = -6. So then you have:

$\lim_{x \rightarrow 0} \frac{6 \cos(2x) + 2ax - 6}{2x}=1$ .

Apply l'Hopital again:

$\lim_{x \rightarrow 0}\frac{-12 \sin (2x) + 2a}{2}=1$ .

In other words, $\frac{2a}{2} = 1 \Rightarrow a = 1$ .

**mr fantastic** · April 21st 2008, 04:15 PM

Originally Posted by akhayoon

Limit question on a midterm that I have and still haven't the slightest idea on how to complete

Find all Values of a and b such that $\lim{x}\rightarrow{0}\frac{3sin(2x)+ax^{2}-bx}{x^2}=1$

all I could figure out was that if you use L'Hopitale's rule you would get

$\lim{x}\rightarrow{0}\frac{6cos(2x)+2ax+b}{2x}=1$

but I have no idea how to use this to solve for any variables

thank you

All your troubles would have been avoided of course if you hadn't run to the l'Hospital with a sore thumb. Try substituting the series expansion around x = 0 for sin(2x):

$\Rightarrow \lim_{x \rightarrow 0} \frac{3 (2x - [2x]^3/6 + ...) + ax^{2} - bx}{x^2}=1$

$\Rightarrow \lim_{x \rightarrow 0} \frac{(6x - 4x^3/3 + ...) + ax^{2} - bx}{x^2}=1$

$\Rightarrow \lim_{x \rightarrow 0} \frac{(6 - b) x + ax^{2} - 4x^3/3 + ...}{x^2}=1$

$\Rightarrow \lim_{x \rightarrow 0} \frac{(6 - b) + ax - 4x^2/3 + ...}{x} = 1$

There's a problem unless 6 - b = 0 => b = 6. Then:

$\Rightarrow \lim_{x \rightarrow 0} \frac{ax - 4x^2/3 + ...}{x} = 1$

$\Rightarrow \lim_{x \rightarrow 0} \left(a - \frac{4x}{3} + ...\right) = 1$

Therefore a = 1.

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